If water vapour is assumed to be perfect gas, molar enthalpy change at 1 bar and 100°C is 41 kJ mol^–1. Calculate the internal energy when 

(i)  1 mol of water is vaporized at 1 bar pressure and 100°C. 

(ii) 1 mol of water is converted into ice.

(i) For vaporization of water, the change is: H₂O(l) → H₂O(g)

Δn_g = 1 − 0 = 1

ΔH = ΔU + Δn_g RT

or    ΔU = ΔH − Δn_g RT = 41.00 kJ mol⁻¹ − (1 mol) × (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)(373 K)

= 41.00 − 3.10 kJ mol⁻¹ = 37.90 kJ mol⁻¹

(ii) For conversion of water into ice, the change is H₂O(l)→H₂O(s)

In this case, the volume change is negligible. Hence ΔH = ΔU 

= 41.00 kJ mol⁻¹.